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Proof step 5 from proof plst-01 (previous step)

i + i0 < j + j0, j0 < j1 ⊦ i0 < i1, i + i1 < j + j1

User applied rule cut formula

And obtained the following new goals

Proof step 6:
j0 < j1, i + i0 < j + j0 ⊦ i0 = i1, i + i1 < j + j1, i0 < i1

Rule argument

fma: i0 = i1

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%lst-01: ⊦ i < j + i0 ∧ j < j0 → i < j0 + i0;% ⊦ ¬ i < i; (s) Used formulas: i0 = i1